Inheritance
Pedigree Charts
Unit 4 A2
People with night blindness have difficulty seeing in dim light. The allele for night
blindness, N, is dominant to the allele for normal vision, n. These alleles are not
carried on the sex chromosomes.
•
•
•
Individual 12 is a boy. What is his phenotype?
What is the genotype of individual 1? Explain the evidence for your answer.
What is the probability that the next child born to individuals 10 and 11 will be a
girl with night blindness? Show your working.
• 1. (a)
Normal sight;
1
• (b) Nn;
Must have at least one N allele as she has the condition
and must pass
on an n allele to her normal sighted children;
2
• (c) Two marks for correct answer of ¼ / 0.25 / 25%;
One mark for incorrect answer that determines
probability of next
child having night blindness as ½ / 0.5 / 50%;
2
max
•
male affected with haemophilia
unaffected male
1
2
Rhesus
positive
3
4
Rhesus
positive
8
Rhesus
negative
Rhesus
positive
unaffected female
Rhesus
negative
5
6
Rhesus
positive
Rhesus
positive
7
Rhesus
positive
9
10
11
Rhesus
positive
Rhesus
positive
Rhesus
positive
12
Rhesus
negative
•
•
•
The allele for Rhesus positive, R, is dominant to that for Rhesus negative, r.
Haemophilia is a sex-linked condition. The allele for haemophilia, h, is recessive to
the allele for normal blood clotting, H, and is carried on the X–chromosome. The
diagram shows the Rhesus blood group phenotypes in a family tree where some
individuals have haemophilia.
Use the information in the diagram to give one piece of evidence that the allele for
the Rhesus negative condition is recessive.
Explain the evidence from the cross between individuals 3 and 4 that the gene
controlling Rhesus blood group is not sex-linked
• a)
• (i) specific cross identified - 3 ,4 & 8 or 10, 11 &
12 /Rh negative
phenotype produced from parents which are
both Rh positive;
1
• (ii) with sex linkage daughter cannot have
(recessive) condition unless
male parent has the condition;
as male passes X chromosome to his daughter;
2
•
1
Affected
male
2
Unaffected
male
3
6
7
4
8
9
Unaffected
female
5
10
11
• Becker muscular dystrophy is a sex linked inherited condition
caused by an allele of a gene on the X chromosome. Sufferers
experience some loss of muscle strength. The diagram shows how
members of one family were affected by the condition.
• Explain one piece of evidence from the diagram which shows that
the allele for Becker muscular dystrophy is recessive.
• The allele for Becker muscular dystrophy is sex-linked. Explain how
individual 9 inherited the condition from his grandfather.
• (b) (i)
3 and 4 do not show the condition but 9/one
male does;
4 must be carrier;
• OR
1 affected but not daughter/4;
who gets X from father; 2
•
• (ii) grandfather/1 passed on his (affected) X chromosome
to his daughter/4;
who was unaffected, because of the ‘normal’ X
inherited from her mother/2;
9 inherited his X chromosome from his mother/4;
2
max
1
5
2
3
6
7
12
Key: Square = male
Circle = female
4
8
9
10
13
14
15
11
16
17
18
red
sandy
white
• The diagram shows the inheritance of coat colour in pigs through three
generations.
• Explain one piece of evidence from the diagram which shows that coat
colour is not controlled by one gene with two codominant alleles.
• sandy stated as heterozygous/suitable allusion to alleles;
suitable cross chosen;(as in table) N.B. second two points linked,
not stand-alone
explained why could not be codominance;
• N.B. Second two points linked, not stand alone
Suitable cross Reason why not codominance
• 3 and 4
Offspring should all be sandy
• 10 and 11 Offspring should all be sandy
• 7 and 8
Offspring should all be red
• BUT if candidate assumes sandy is homozygous, mark accordingly
e.g. “look at cross 1 and 2; all their offspring would be sandy;”
and not that, if red or white then identified as heterozygote,
then full 3 marks are still possible.
Affected
male
1
2
Unaffected
male
Affected
female
3
5
•
•
•
•
6
7
4
8
9
Unaffected
female
10
Nail-patella syndrome is an inherited condition caused by a single gene. Sufferers
have abnormal nail growth and underdeveloped kneecaps. The pedigree shows
how members of one family were affected by the syndrome.
Explain one piece of evidence from the pedigree which indicates that
(i)
the allele for the nail-patella syndrome is dominant;
(ii) the gene is not sex-linked.
• (b) (i) 3 and 4 produce unaffected male/8 /
female/10, so must
3 must be a carrier recessive;
if the condition is expressed in the
heterozygous form, then the condition must
be dominant 2
• 4 would inherit an X chromosome form her
father which would carry the dominant allele
and thus she would express the condition, as
she does not it can not be on the X
•
•
•
•
The Rhesus blood group is genetically controlled. The gene for the Rhesus blood
group has two alleles. The allele for Rhesus positive, R, is dominant to that for
Rhesus negative, r. The diagram shows the inheritance of the Rhesus blood group
in one family.
Explain one piece of evidence from the diagram which shows that the allele for
Rhesus positive is dominant.
Explain one piece of evidence from the diagram which shows that the gene is not
on the X chromosome.
Sixteen percent of the population of Europe is Rhesus negative. Use the HardyWeinberg equation to calculate the percentage of this population that you would
expect to be heterozygous for the Rhesus gene. Show your working. (3)
• 3 and 4 / two Rhesus positives produce Rhesus negative child/children / 7
/ 9;
• Both Rhesus positives/3 and 4 carry recessive (allele)/ are heterozygous /
if Rhesus positive was recessive, all children (of 3 and 4) would be Rhesus
positive/recessive;
• 3 would not be/is Rhesus positive / would be Rhesus negative;
• 3 would receive Rhesus negative (allele) on X (chromosome) from mother
/ 3 could not receive Rhesus positive (allele) from mother / 3 would not
receive Rhesus positive (allele)/X (chromosome) from father/1 / 3 will
receive Y (chromosome) from father/1;
• OR
• 9 would be Rhesus positive / would not be/is Rhesus negative / 8 and 9/all
daughters of 3 and 4 would be Rhesus positive;
• As 9 would receive X chromosome/dominant allele from father/3;
• Correct answer of 48(%) = 3 marks;;;
• q2/ p2= 16%/0.16 / p/q = 0.4;
• Shows that 2pq = heterozygotes / carriers;
• A single gene controls the presence of hair on the skin of
cattle. The gene is carried on the X chromosome. Its
dominant allele causes hair to be present on the skin and
its recessive allele causes hairlessness. The diagram shows
the pattern of inheritance of these alleles in a group of
cattle.
• Use evidence from the diagram to explain
• that hairlessness is caused by a recessive allele
• that hairlessness is caused by a gene on the X chromosome.
•
•
•
•
•
•
1. Animal 2 / 5 has hair but offspring do not;
2. So 2 / 5 parents must be heterozygous/carriers;
OR
3. 4/7/8 are hairless but parents have hair;
4. So 2 / 5 must be heterozygous/carriers;
Accept parents as alternative to animals 2 and 5,
1 + 3: Allow reference to children/offspring for
animals 7 + 8
• Hairless males have fathers with hair / 4 is
hairless but 1 is hairy / 7 and/or 8 are hairless but
6 is hairy / only males are hairless
•
•
•
•
•
Tay-Sachs disease is a human inherited disorder. Sufferers of this disease often die
during childhood. The allele for Tay-Sachs disease t, is recessive to allele T, present
in unaffected individuals. The diagram shows the inheritance of Tay-Sachs in one
family.
Explain one piece of evidence from the diagram which proves that the allele for
Tay-Sachs disease is recessive.
Explain one piece of evidence from the diagram which proves that the allele for
Tay-Sachs disease is not on the X chromosome.
In a human population, one in every 1000 children born had Tay-Sachs disease.
Use the Hardy-Weinberg equation to calculate the percentage of this population
you would expect to be heterozygous for this gene. Show your working.(3)
• 1. 3 and 4 and 9/11/affected offspring;
• 2. Both 3 and 4 are carriers/heterozygous;
OR
• If dominant at least one of 3 and 4 would be
affected;
• 1 Accept: 9/11 and their parents
• 1 Accept: unaffected parents have affected
children
• 2 Accept: if 3 and 4 are unaffected all their
children will be unaffected
•
•
• 1. 11 is affected, 3 is not;
• 2. 3/father of 11 does not have a recessive allele on his X
chromosome/ Xt;
• OR
• (If on X) 11/affected female would not receive the recessive
allele on X chromosome/Xt from 3/father;
• OR
• (If on X) 3/father (of 11) would pass on the dominant allele
on his X chromosome/XT;
•
•
•
•
•
Answer in range of 5.8 - 6.2% = 3 marks;;;
If incorrect answer, then 2 max of following points
1. q2/p2/tt = 0.001 or 1 divided by 1000;
2. p/q/T = 0.968 – 0.97;
3. Understanding that heterozygous = 2pq;